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Question: A point P (t<sup>2</sup>, 2t) lies on the parabola y<sup>2</sup> = 4x, where FP is produced to B whe...

A point P (t2, 2t) lies on the parabola y2 = 4x, where FP is produced to B where F is focus. If PB = ƒ(t) and point B always lies on the line y – x = 2, then ƒ(t) is equal to-

A

(t2−2t+2)(1+t2)−t2+2t+1\frac{(t^{2} - 2t + 2)(1 + t^{2})}{- t^{2} + 2t + 1}

B

(t2−2t+2)(1+t2)t2+2t+1\frac{(t^{2} - 2t + 2)(1 + t^{2})}{t^{2} + 2t + 1}

C

(t2−2t+2)(1−t2)−t2+2t+1\frac{(t^{2} - 2t + 2)(1 - t^{2})}{- t^{2} + 2t + 1}

D

None of these

Answer

(t2−2t+2)(1+t2)−t2+2t+1\frac{(t^{2} - 2t + 2)(1 + t^{2})}{- t^{2} + 2t + 1}

Explanation

Solution

Let q be the angle which the line FP makes with positive direction of x-axis, then

tan q = 2tt2−1\frac{2t}{t^{2} - 1}

sin q = 2t1+t2\frac{2t}{1 + t^{2}}, cos q = t2−11+t2\frac{t^{2} - 1}{1 + t^{2}}

B ŗ (1+(1+t2+ƒ(t))t2−11+t2,(1+t2+ƒ(t))2t1+t2)\left( 1 + (1 + t^{2} + ƒ(t))\frac{t^{2} - 1}{1 + t^{2}},(1 + t^{2} + ƒ(t))\frac{2t}{1 + t^{2}} \right)

\ B = lies on y – x = 2

Ž (1 + t2 + ƒ(t)) 2t1+t2\frac{2t}{1 + t^{2}}

= 1 + (1 + t2 + ƒ(t)) t2−11+t2+2\frac{t^{2} - 1}{1 + t^{2}} + 2

Ž ƒ(t) (2t−t2+11+t2)\left( \frac{2t - t^{2} + 1}{1 + t^{2}} \right)= 3 + t2 – 1 – 2t

Ž ƒ(t) = (t2−2t+2)(1+t2)−t2+2t+1\frac{(t^{2} - 2t + 2)(1 + t^{2})}{- t^{2} + 2t + 1}

Hence (1) is correct answer.